Question

Let $\alpha ,\beta$ are the angle of inclination of the tangents to the axis of the parabola $y^2=4ax$ drawn from the point $P.$

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}\cot \alpha \cot \beta =k,\text{then locus of}P\text{is}& \text{(P)}& kx=a\\ \text{(B)}& \text{If}\tan \alpha +\tan \beta =k,\text{then locus of}P\text{is}& \text{(Q)}& y=k(x-a)\\ \text{(C)}& \text{If}\tan (\alpha +\beta )=k,\text{then locus of}P\text{is}& \text{(R)}& kx=y\\ \text{(D)}& \text{If}\tan \alpha \tan \beta =k,\text{then locus of}P\text{is}& \text{(S)}& xy=k\\ & & \text{(T)}& x=ka\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(A)}\to \text{(T)},\text{(B)}\to \text{(R)},\text{(C}\to \text{(Q)},\text{(D)}\to \text{(P)}$
• B. $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(Q)},\text{(C)}\to \text{(P)},\text{(D)}\to \text{(T)}$
• C. $\text{(A)}\to \text{(P)},\text{(B)}\to \text{(Q)},\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(S)}$
• D. $\text{(A)}\to \text{(S)},\text{(B)}\to \text{(Q)},\text{(C)}\to \text{(P)},\text{(D)}\to \text{(T)}$

Answer 1

The correct option is A $\text{(A)}\to \text{(T)},\text{(B)}\to \text{(R)},\text{(C}\to \text{(Q)},\text{(D)}\to \text{(P)}$

Equation of tangent to parabola $y^2=4ax$ is :
$y=mx+\frac{a}{m}\Rightarrow m^{2}x-my+a=0$
As it passes through $P(x_1,y_1),$ then
$\Rightarrow m^2{x}_1-m{y}_1+a=0$

Given, $\alpha ,\beta$ are the angles made by the tangents with the $x-$ axis, then
$\Rightarrow \tan \alpha +\tan \beta =\frac{y_1}{x_1}$
and $\tan \alpha \tan \beta =\frac{a}{x_1}$

$\left(A\right)\cot \alpha \cot \beta =k$
$\Rightarrow \tan \alpha \tan \beta =\frac{1}{k}=\frac{a}{x_1}$
Required locus is $x=ka.$

$\left(B\right)\tan \alpha +\tan \beta =k=\frac{y_1}{x_1}$
Required locus is $y=kx.$

$\left(C\right)\tan (\alpha +\beta )=k$
$\Rightarrow \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=k\Rightarrow \frac{\frac{y_1}{x_1}}{1-\frac{a}{x_1}}=k$
Required locus is $y=k(x-a).$

$\left(D\right)\tan \alpha \tan \beta =k=\frac{a}{x_1}$
Required locus is $kx=a.$

$\therefore \text{(A)}\to \text{(T)},\text{(B)}\to \text{(R)},\text{(B)}\to \text{(Q)},\text{(D)}\to \text{(P)}$